Question

question 1 (5 points)
if p, q, and r are any three collinear points such that q lies between p and r; find the value of x if pr = 5x, pq = 10, and qr = 3x - 2.
1
8
6
4
question 2 (5 points)
find the distance between the points x = - 7 and y = 3 using the ruler postulate.
4
10
-4
-10

Explanation:

Step1: Use segment - addition postulate for Question 1

Since \(P\), \(Q\), and \(R\) are collinear and \(Q\) lies between \(P\) and \(R\), then \(PR = PQ+QR\).
Substitute \(PR = 5x\), \(PQ = 10\), and \(QR=3x - 2\) into the equation: \(5x=10+(3x - 2)\).

Step2: Simplify the equation for Question 1

First, simplify the right - hand side: \(5x=10 + 3x-2\), which becomes \(5x=3x + 8\).
Then, subtract \(3x\) from both sides: \(5x-3x=3x + 8-3x\), so \(2x=8\).
Finally, divide both sides by \(2\): \(x = 4\).

Step1: Use the ruler postulate for Question 2

The ruler postulate states that the distance \(d\) between two points \(a\) and \(b\) on a number line is \(d=\vert a - b\vert\).
Here, \(a=-7\) and \(b = 3\), so \(d=\vert-7 - 3\vert\) or \(d=\vert3-(-7)\vert\).

Step2: Calculate the absolute value for Question 2

\(\vert-7 - 3\vert=\vert-10\vert = 10\) and \(\vert3-(-7)\vert=\vert3 + 7\vert=\vert10\vert = 10\).

Answer:

1. \(8\)
2. \(10\)